Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Deriving Posseuille's Law

  1. Jan 6, 2017 #1
    Hello,

    I’ve been trying to derive Posseuille’s Law and I’m very close. However there seems to be a small difference between my concluded formula and the real one but I don’t know why.

    Here’s what I get:

    Picture a cilindrical tube with a radius ##R## in which a fluid is flowing. From what I know, there are 2 forces that play a role here. There’s the force ##F_P## that causes the flow itself and is proportional to the pressure difference ##ΔP##; it’s equal to ##F_P = ΔP \cdot π \cdot R^2##. And there’s a force ##F_V## that counteracts ##F_P## because of the fluid’s viscosity ## ɳ ## which causes friction at the walls of the tube; it’s equal to ##F_V = ɳ \cdot 2 \cdot R \cdot L \cdot \frac{v}{r}##. The ##r## is a chosen radius where you want to know ## F_V ##.

    From what I understand, the ##F_P## is equal to ##F_V## at every chosen radius ##r## within radius ##R## of the tube. So we can say that:

    $$ \frac{ΔP \cdot π \cdot R^2}{ɳ \cdot 2 \cdot \cdot π \cdot R \cdot L} = \frac{v}{r} $$

    To calculate ##v## here, one would have to rewrite this as:

    $$ \frac{ΔP \cdot R \cdot r}{ɳ \cdot 2 \cdot L} = v $$

    We here now have written ##v## as a function of a chosen ##r## within the tube. If we now choose very small ##r##’s, we would get more or less the velocity of the portion of the fluid within that chosen ##r##. That velocity times the surface ##π \cdot r^2## would give the flow rate ##Q## through that chosen ##r##. We’d thus have to integrate the formula:

    $$ \frac{ΔP \cdot R \cdot r}{ɳ \cdot 2 \cdot L} \cdot π \cdot r^2 = \frac{ΔP \cdot R \cdot r^3 \cdot π}{ɳ \cdot 2 \cdot L} $$

    Integrating this would finally give the formula:

    $$ \frac{ΔP \cdot R \cdot π \cdot r^4}{8 \cdot L \cdot ɳ} = Q $$

    The correct formula is missing the ##R## parameter in the numerator, which is the radius of the tube. I don’t get why that is missing since that’s the determining factor in the formulae for the ##F_P## and ##F_V##.
     
    Last edited: Jan 6, 2017
  2. jcsd
  3. Jan 6, 2017 #2
    Where did you get your equation for Fv from? To me, it makes no sense.
     
  4. Jan 6, 2017 #3
    Got it from here: http://www.insula.com.au/physics/1250/poiseuille.html

    Looking at it now, I noticed it doesn't fit since a smaller ##r## should give a smaller ##F_V##. How should it be noted then?
     
  5. Jan 6, 2017 #4
    If should be $$F_v=(2\pi R L)\left(-\eta \frac{dv}{dr}\right)_{r=R}$$And dv/dr is not constant. v is maximum at the center of the tube and zero at the wall.
     
  6. Jan 6, 2017 #5
    I don't really get what ##dr## is. Is it ##R - r##? And how about ##dv##? How would one get the actual velocity at a particular ##r## with that formula?
     
  7. Jan 6, 2017 #6
    It's hard to know where to start without knowing more about your background. Have you had calculus yet?
     
  8. Jan 6, 2017 #7
    It was a long time ago for me during highschool. However, after watching a video, I conclude that ##dv## is the difference of velocity between a radius of 0 and a chosen ##r## ( = ##dr##), inside the tube. This ##dv## should be negative since increasing the ##dr## lowers the velocity because it approaches the walls of the tube, hence the minus sign in the formula. Because the velocity at the walls is 0, I can calculate the velocity at the middle of the tube if I use ##dr## as ##R##, thus:

    $$ v_(max) = \frac{ΔP \cdot R^2}{ɳ \cdot 2 \cdot L} $$

    Now that I have the maximum velocity at the middle of the tube (##dr = 0##), I can calculate the velocity at any ##dr## away from ##dr = 0## by substracting your mentioned formula with any chosen ##dr##, from my concluded formula. Thus:

    $$ v = \frac{ΔP \cdot R^2}{ɳ \cdot 2 \cdot L} - \frac{ΔP \cdot dr^2}{ɳ \cdot 2 \cdot L} = \frac{ΔP}{ɳ \cdot 2 \cdot L}(R^2 - dr^2) $$

    Am I going the right way here? If I am, then this is the formula of ##v(r)##. Now, if I choose a very small ##r##, calculate ##v## out of that and then multiply this ##v## with the surface of that very small ##r## which is ##π \cdot r^2##, I'd get the volume rateflow within that small surface. I'd have to add all these small surfaces until I reach a radius of ##R##. Thus I'd have to integrate:

    $$ \frac{ΔP}{ɳ \cdot 2 \cdot L}(R^2 - dr^2) \cdot π \cdot dr^2 $$

    Which still doesn't give me the correct equation for volume rate flow. What am I doing wrong here? Should I substract each small surface from the previous surface in the formula?
     
    Last edited: Jan 6, 2017
  9. Jan 7, 2017 #8
    None of this is correct. You are not going to be able to do this without using calculus, and it looks like your calculus skills are currently not adequate.

    For example: dv is the difference between the velocity at locations r and r + dr, and dr is not the radial location; it is the tiny incremental difference between two closely neighboring radial locations.
     
  10. Jan 7, 2017 #9
    Ok, I do remember some calculus but I thought that it could be interpreted in another way. I think I have 2 questions that would make things more clear for me.

    1. If ##\frac{dv}{dr}## is the derivative of a function ##v(r)##, then doesn't this mean that ##\frac{ΔP \cdot r}{ɳ \cdot 2 \cdot L}## is the derivative as well of the function ##v(r)##? So I'd have to integrate that formula to get this function?

    2. You say that ##dr## is the difference in velocity between r and r + dr. If I set ##r## to 0 and choose ##dr## to be ##R##, would the formula give me the difference of ##v## between a radius of ##R## and a radius of 0?
     
  11. Jan 7, 2017 #10
    No, no, and no. If you want to see the correct derivation using a "shell" force balance, see Transport Phenomena by Bird, Stewart, and Lightfoot.
     
  12. Jan 7, 2017 #11
    Then why are there people who are integrating that formula as part of deriving Posseuille's Law?

    Image470.gif
    Source: http://www.insula.com.au/physics/1250/poiseuille.html

    Also, this lecturer does the same at 5:15:

     
  13. Jan 7, 2017 #12
  14. Jan 7, 2017 #13
    But that's what I proposed in my first question, integrating ##\frac{ΔP \cdot r}{ɳ \cdot 2 \cdot L}## to get the function of ##v(r)##. Why wasn't it correct?
     
  15. Jan 7, 2017 #14
    You are aware that there is a difference between dv/dr and v/r, correct?
     
  16. Jan 10, 2017 #15
    Yes, I am after your explanation in your post #8. The only thing that bothers me about it though is why, when you choose r = 0 and r + dr = R (thus ##dr = R##), it doesn't give the difference between the velocity (##dv##) in the middle of the tube (##r = 0##) and the velocity at radius of the tube (##R##). Since the velocity at ##R## is 0, the difference should be equal to the velocity at the middle of the tube, which is the maximum velocity in the tube. This may be all out of the subject of deriving Poisseulle's law but I'd really like to understand this first. Is there a way to explain this?
     
  17. Jan 10, 2017 #16
    There is a key difference between dr and ##\Delta r##, and between dv and ##\Delta v##. dr is strictly reserved for tiny differences in r, and dv is reserved for tiny differences in v. It is improper to write dr = R - 0.

    dv/dr can only be set equal to v/r if we know for a fact that v is directly proportional to r. But for flow in a tube, v is not directly proportional to r. For flow in a tube, v varies quadratically with r. Look at the final equation for v as a function of r. So you can't just set ##dv/dr =-v_{max}/R##. It is incorrect mathematically.
     
  18. Jan 10, 2017 #17

    boneh3ad

    User Avatar
    Science Advisor
    Gold Member

    Perhaps it is fruitful here to discuss the definition of a derivative in this case?

    [tex]\dfrac{dv}{dr} = \lim_{\Delta r \to 0}\dfrac{v(r+\Delta r) - v(r)}{\Delta r}[/tex]

    So if you just pick arbitrary values for ##r## for ##\Delta r##, you aren't satisfying the limit. Sometimes you can approximate a derivative by doing something similar to that (see: finite differences), but it isn't even approximately correct when you take your ##\Delta r## to be the entire domain.
     
  19. Jan 16, 2017 #18
    That's where my confusion was! I didn’t know there was a difference between ##Δr## and ##dr##. I do understand that choosing a fixed ##Δr## would give a different ##Δv## since the relation is a parabola and thus ##Δv## would depend on which r to which r you want to take that fixed ##Δr## from. But apparently ##\frac{dv}{dr}## really is about a very local steepness of a curve instead of a larger range.

    So, after integration I’d get the following function of ##v(r)##:

    $$v(r) = \frac{ΔP}{4ηL} \cdot (R^2-r^2)$$

    Now, to get the flow per time unit I’d have to multiply that formula ##v(r)## with the area of the tube ##π \cdot r^2##. But since the velocity differs with ##r## I’d have to integrate the whole formula over small ##dr##’s.

    Here’s where I’m stuck, because if I choose to integrate the circle area ##π \cdot dr^2##, then the integration would just keep integrating a fixed circle area. That’s incorrect because substracting a circle area of ##π \cdot (n \cdot dr)^2## from a circle area of ##π \cdot ((n-1) \cdot dr)^2)## with ##n## being whole numbers, gives circle areas of ##πr^2##, ##3 \cdot πr^2##, ##5 \cdot πr^2##, etc. I don’t know how to put this in a formula before starting the integration.

    Watching the previous video I gave a link to shows that I’d have to integrate ##2 \cdot π \cdot r \cdot dr## multiplied by ##v(r)##. But I don’t have the slightest idea how he came to ##2 \cdot π \cdot r \cdot dr## based on the following picture:

    Tube.jpg
     
    Last edited: Jan 16, 2017
  20. Jan 21, 2017 #19
    Anyone who could please help me with my question in the above post?
     
  21. Jan 21, 2017 #20
    The area over which v(r) applies is the anulus of area ##\pi(r+\frac{dr}{2})^2-\pi (r-\frac{dr}{2})^2##, where r is the radial location at the mid-point of the anulus. But this is just equal to ##2\pi r dr##.

    Also, from calculus, ##d(r^2)=2rdr##, so, in your own development, ##\pi d(r^2)=2\pi r dr##
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Deriving Posseuille's Law
  1. Derive Ohm's law? (Replies: 16)

Loading...